# R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 10 - Quadrilaterals

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## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 10 - Quadrilaterals Page/Excercise MCQ

Correct option: (a)

ABCD is a rhombus.

⇒ AD ∥ BC and AC is the transversal.

⇒ ∠DAC = ∠ACB (alternate angles)

⇒ ∠DAC = 50°

In ΔAOD, by angle sum property,

∠AOD + ∠DAO + ∠ADO = 180°

⇒ 90° + ∠50° + ∠ADO = 180°

⇒ ∠ADO = 40°

⇒ ∠ADB = 40°

Correct option: (b)

∠DAO + ∠OAB = ∠DAB

⇒ ∠DAO + 35° = 90°

⇒ ∠DAO = 55°

ABCD is a rectangle and diagonals of a rectangle are equal and bisect each other.

OA = OD

⇒ ∠ODA = ∠DAO (angles opposte to equal sides are equal)

⇒ ∠ODA = 55°

In DODA, by angle sum property,

∠ODA + ∠DAO + ∠AOD = 180°

⇒ 55° + ∠55° + ∠AOD = 180°

⇒ ∠AOD = 70°

Correct option: (d)

In ΔAPB, by angle sum property,

∠APB + ∠PAB + ∠PBA = 180°

In ΔCRD, by angle sum property,

∠CRD + ∠RDC + ∠RCD = 180°

Now, ∠SPQ + ∠SRQ = ∠APB + ∠CRD

= 360° - 180°

= 180°

Now, ∠PSR + ∠PQR = 360° - (∠SPQ + ∠SRQ)

= 360° - 180°

= 180°

Hence, PQRS is a quadrilateral whose opposite angles are supplementary.

Correct option: (d)

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii),

PQ ∥ RS and PQ = RS

Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So, PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

Now, in ΔABD, P and S are the mid-points of sides AB and AD respectively.

Thus, in quadrilateral PMON, PM ∥ NO and PN ∥ MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90°

Thus, PQRS is a parallelogram whose one angle, i.e. ∠QPS = 90°.

Hence, PQRS is a rectangle if AC ⊥ BD.

Correct option: (c)

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ ∥ RS and QR ∥ SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Hence, PQRS is a rhombus if diagonals of ABCD are equal.

Correct option: (d)

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Let the diagonals AC and BD intersect at O.

Now,

Thus, in quadrilateral PMON, PM || NO and PN || MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90°

Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and ∠QPS = 90°.

Hence, PQRS is a square if diagonals of ABCD are equal and perpendicular.

Correct option: (b)

In ΔABC, D and E are the mid-points of sides AB and AC respectively.

Hence, DENM is a parallelogram.

Since opposite angles of quadrilateral are equal, PQRS is a parallelogram.

⇒ PQ = SR (opposite sides of parallelogram are equal)

⇒ PQ = 2 cm

The given statement is false.

Diagonals of a parallelogram bisect each other.

In quadrilateral PQRS, ∠P and ∠S are adjacent angles.

Since the sum of adjacent angles ≠ 180°, PQRS is not a parallelogram.

Hence, PQRS is a trapezium.

The given statement is false.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are acute, the sum will be less than 360°.

The given statement is true.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are right angles,

Sum of all angles of a quadrilateral = 4 × 90° = 360°

The given statement is false.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are obtuse, the sum will be more than 360°.

We know that the sum of all the four angles of a quadrilateral is 360°.

Here,

70° + 115° + 60° + 120° = 365° ≠ 360°

Hence, we cannot form a quadrilateral with given angles.

A quadrilateral whose all angles are equal is a rectangle.

D and E are respectively the midpoints of the sides AB and BC of ΔABC.

Thus, by mid-point theorem, we have

Since the diagonals PR and QS of quadrilateral PQRS bisect each, PQRS is a parallelogram.

Now, adjacent angles of parallelogram are supplementary.

⇒ ∠Q + ∠R = 180°

⇒ 56° + ∠R = 180°

⇒ ∠R = 124°

AFDE is a parallelogram

⇒ AF = ED …(i)

BDEF is a parallelogram.

⇒ FB = ED …(ii)

From (i) and (ii),

AF = FB

Correct option: (a)

If the diagonals of a quad. ABCD bisect each other, then the quad. ABCD is a parallelogram.

So, I gives the answer.

If the diagonals are equal, then the quad. ABCD is a parallelogram.

So, II gives the answer.

Correct option: (c)

If the quad. ABCD is a ‖gm, it could be a rectangle or square or rhombus.

So, statement I is not sufficient to answer the question.

If the diagonals AC and BD are perpendicular to each other, then the ‖gm could be a square or rhombus.

So, statement II is not sufficient to answer the question.

However, if the statements are combined, then the quad. ABCD is a rhombus.

Correct option: (c)

If the diagonals of a ‖gm ABCD are equal, then ‖gm ABCD could either be a rectangle or a square.

If the diagonals of the ‖gm ABCD intersect at right angles, then the ‖gm ABCD could be a square or a rhombus.

However, if both the statements are combined, then ‖gm ABCD will be a square.

Correct option: (b)

If the opposite sides of a quad. ABCD are equal, the quadrilateral is a parallelogram.

If the opposite angles are equal, then the quad. ABCD is a parallelogram.

The Reason (R) is true and is the correct explanation for the Assertion (A).

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 10 - Quadrilaterals Page/Excercise 10A

Let the measure of the fourth angle = x°

For a quadrilateral, sum of four angles = 360°

⇒ x° + 75° + 90° + 75° = 360°

⇒ x° = 360° - 240°

⇒ x° = 120°

Hence, the measure of fourth angle is 120°.

Since AB || DC

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Given: O is a point within a quadrilateral ABCD

Given: ABCD is a quadrilateral and AC is one of its disgonals.

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Given: ABCD is a quadrilateral.

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## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 10 - Quadrilaterals Page/Excercise 10B

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ABCD is a parallelogram.

Hence, AD || BC.

⇒ ∠DAM = ∠AMB (alternate angles)

⇒ ∠BAM = ∠AMB (since ∠BAM = ∠DAM)

⇒ BM = AB (sides opposite to equal angles are equal)

But, AB = CD (opposite sides of a parallelogram)

⇒ BM = AB = CD ….(i)

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Let the altitude from D to the side AB bisect AB at point P.

Join BD.

In ΔAMD and ΔBMD,

AM = BM (M is the mid-point of AB)

∠AMD = ∠BMD (Each 90°)

MD = MD (common)

∴ ΔAMD ≅ ΔBMD (by SAS congruence criterion)

⇒ AD = BD (c.p.c.t.)

But, AD = AB (sides of a rhombus)

⇒ AD = AB = BD

⇒ ΔADB is an equilateral triangle.

⇒ ∠A = 60°

⇒ ∠C = ∠A = 60° (opposite angles are equal)

⇒ ∠B = 180° - ∠A = 180° - 60° = 120°

⇒ ∠D = ∠B = 120°

Hence, in rhombus ABCD, ∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°.

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In ΔABC and ΔADC,

AB = AD (sides of a rhombus are equal)

BC = CD (sides of a rhombus are equal)

AC = AC (common)

∴ ΔABC ≅ ΔADC (by SSS congruence criterion)

⇒ ∠BAC = ∠DAC and ∠BCA = ∠DCA (c.p.c.t.)

⇒ AC bisects ∠A as well as ∠C.

Similarly,

In ΔBAD and ΔBCD,

AB = BC (sides of a rhombus are equal)

AD = CD (sides of a rhombus are equal)

BD = BD (common)

∴ ΔBAD ≅ ΔBCD (by SSS congruence criterion)

⇒ ∠ABD = ∠CBD and ∠ADB = ∠CDB (c.p.c.t.)

⇒ BD bisects ∠B as well as ∠D.

In ΔAMO and ΔCNO

∠MAO = ∠NCO (AB ∥ CD, alternate angles)

AM = CN (given)

∠AOM = ∠CON (vertically opposite angles)

∴ ΔAMO ≅ ΔCNO (by ASA congruence criterion)

⇒ AO = CO and MO = NO (c.p.c.t.)

⇒ AC and MN bisect each other.

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∠DCM = ∠DCN + ∠MCN

⇒ 90° = ∠DCN + 60°

⇒ ∠DCN = 30°

In ΔDCN,

∠DNC + ∠DCN + ∠D = 180°

⇒ 90° + 30° + ∠D = 180°

⇒ ∠D = 60°

⇒ ∠B = ∠D = 60° (opposite angles of parallelogram are equal)

⇒ ∠A = 180° - ∠B = 180° - 60° = 120°

⇒ ∠C = ∠A = 120°

Thus, the angles of a parallelogram are 60°, 120°, 60° and 120°.

(i) ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

⇒ ∠BAC = ∠DAC ….(i)

And ∠BCA = ∠DCA ….(ii)

Since every rectangle is a parallelogram, therefore

AB ∥ DC and AC is the transversal.

⇒ ∠BAC = ∠DCA (alternate angles)

⇒ ∠DAC = ∠DCA [From (i)]

Thus, in ΔADC,

AD = CD (opposite sides of equal angles are equal)

But, AD = BC and CD = AB (ABCD is a rectangle)

⇒ AB = BC = CD = AD

Hence, ABCD is a square.

(ii) In ΔBAD and ΔBCD,

AB = CD

AD = BC

BD = BD

∴ ΔBAD ≅ ΔBCD (by SSS congruence criterion)

⇒ ∠ABD = ∠CBD and ∠ADB = ∠CDB (c.p.c.t.)

Hence, diagonal BD bisects ∠B as well as ∠D.

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l ∥ m and t is a transversal.

⇒ ∠APR = ∠PRD (alternate angles)

⇒ ∠SPR = ∠PRQ (PS and RQ are the bisectors of ∠APR and ∠PRD)

Thus, PR intersects PS and RQ at P and R respectively such that ∠SPR = ∠PRQ i.e., alternate angles are equal.

⇒ PS ∥ RQ

Similarly, we have SR ∥ PQ.

Hence, PQRS is a parallelogram.

Now, ∠BPR + ∠PRD = 180° (interior angles are supplementary)

⇒ 2∠QPR + 2∠QRP = 180° (PQ and RQ are the bisectors of ∠BPR and ∠PRD)

⇒ ∠QPR + ∠QRP = 90°

In ΔPQR, by angle sum property,

∠PQR + ∠QPR + ∠QRP = 180°

⇒ ∠PQR + 90° = 180°

⇒ ∠PQR = 90°

Since PQRS is a parallelogram,

∠PQR = ∠PSR

⇒ ∠PSR = 90°

Now, ∠SPQ + ∠PQR = 180° (adjacent angles in a parallelogram are supplementary)

⇒ ∠SPQ + 90° = 180°

⇒ ∠SPQ = 90°

⇒ ∠SRQ = 90°

Thus, all the interior angles of quadrilateral PQRS are right angles.

Hence, PQRS is a rectangle.

AK = BL = CM = DN (given)

⇒ BK = CL = DM = AN (i)(since ABCD is a square)

In ΔAKN and ΔBLK,

AK = BL (given)

∠A = ∠B (Each 90°)

AN = BK [From (i)]

∴ ΔAKN ≅ ΔBLK (by SAS congruence criterion)

⇒ ∠AKN = ∠BLK and ∠ANK = ∠BKL (c.p.c.t.)

But, ∠AKN + ∠ANK = 90° and ∠BLK + ∠BKL = 90°

⇒ ∠AKN + ∠ANK + ∠BLK + ∠BKL = 90° + 90°

⇒ 2∠AKN + 2∠BKL = 180°

⇒ ∠AKN + ∠BKL = 90°

⇒ ∠NKL = 90°

Similarly, we have

∠KLM = ∠LMN = ∠MNK = 90°

Hence, KLMN is a square.

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## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 10 - Quadrilaterals Page/Excercise 10C

(i) In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

(ii) In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii), we have

PQ = SR and PQ ∥ SR

(iii) Thus, in quadrilateral PQRS, one pair of opposite sides are equal and parallel.

Hence, PQRS is a parallelogram.

Let ΔABC be an isosceles right triangle, right-angled at B.

⇒ AB = BC

Let PBSR be a square inscribed in ΔABC with common ∠B.

⇒ PB = BS = SR = RP

Now, AB - PB = BC - BS

⇒ AP = CS ….(i)

In ΔAPR and ΔCSR

AP = CS [From (i)

∠APR = ∠CSR (Each 90°)

PR = SR (sides of a square)

∴ ΔAPR ≅ ΔCSR (by SAS congruence criterion)

⇒ AR = CR (c.p.c.t.)

Thus, point R bisects the hypotenuse AC.

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In ΔAOM and ΔCON

∠MAO = ∠OCN (Alternate angles)

AO = OC (Diagonals of a parallelogram bisect each other)

∠AOM = ∠CON (Vertically opposite angles)

∴ ΔAOM ≅ ΔCON (by ASA congruence criterion)

⇒ MO = NO (c.p.c.t.)

Thus, MN is bisected at point O.

Construction: Join diagonal QS. Let QS intersect MN at point O.

PQ ∥ SR and MN ∥ PQ

⇒ PQ ∥ MN ∥ SR

By converse of mid-point theorem a line drawn, through the mid-point of any side of a triangle and parallel to another side bisects the third side.

Now, in ΔSPQ

MO ∥ PQ and M is the mid-point of SP

So, this line will intersect QS at point O and O will be the mid-point of QS.

Also, MN ∥ SR

Thus, in ΔQRS, ON ∥ SR and O is the midpoint of line QS.

So, by using converse of mid-point theorem, N is the mid-point of QR.

PM is the bisector of ∠P.

⇒ ∠QPM = ∠SPM ….(i)

PQRS is a parallelogram.

∴ PQ ∥ SR and PM is the transversal.

⇒ ∠QPM = ∠MS (ii)(alternate angles)

From (i) and (ii),

∠SPM = ∠PMS ….(iii)

⇒ MS = PS = 9 cm (sides opposite to equal angles are equal)

Now, ∠RMT = ∠PMS (iv)(vertically opposite angles)

Also, PS ∥ QT and PT is the transversal.

∠RTM = ∠SPM

⇒ ∠RTM = ∠RMT

⇒ RT = RM (sides opposite to equal angles are equal)

RM = SR - MS = 12 - 9 = 3 cm

⇒ RT = 3 cm

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In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Hence, PQRS is a rhombus.

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii),

PQ || RS and PQ = RS

Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So, PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

Now, in ΔABD, P and S are the mid-points of sides AB and AD respectively.

Thus, in quadrilateral PMON, PM || NO and PN || MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90°

Thus, PQRS is a parallelogram whose one angle, i.e. ∠QPS = 90°.

Hence, PQRS is a rectangle.

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Let the diagonals AC and BD intersect at O.

Now,

Thus, in quadrilateral PMON, PM || NO and PN || MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90°

Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and ∠QPS = 90°.

Hence, PQRS is a square.

## R. S. Aggarwal and V. Aggarwal - Mathematics - IX Class 9 Chapter Solutions

- Chapter 1 - Number Systems
- Chapter 2 - Polynomials
- Chapter 3 - Factorisation of Polynomials
- Chapter 4 - Linear Equations in Two Variables
- Chapter 5 - Coordinate Geometry
- Chapter 6 - Introduction to Euclid's Geometry
- Chapter 7 - Lines And Angles
- Chapter 8 - Triangles
- Chapter 9 - Congruence of Triangles and Inequalities in a Triangle
- Chapter 10 - Quadrilaterals
- Chapter 11 - Areas of Parallelograms and Triangles
- Chapter 12 - Circles
- Chapter 13 - Geometrical Constructions
- Chapter 14 - Areas of Triangles and Quadrilaterals
- Chapter 15 - Volume and Surface Area of Solids
- Chapter 16 - Presentation of Data in Tabular Form
- Chapter 17 - Bar Graph, Histogram and Frequency Polygon
- Chapter 18 - Mean, Median and Mode of Ungrouped Data
- Chapter 19 - Probability

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